Question

A coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperature 80.000 C. A salt weighing 7.093 g is quickly added. The salt has a molar mass of 379.984 g/mol. The final temperature of the solution is 61.128 C. Assume no heat loss to the surroundings. Assume the specific heat capacity of the solution is equal to that of pure water, and that the mass of the solution is equal to the mass of the solid plus the mass of water in the calorimeter. What is the molar heat of solution for the salt, in kJ/mol? Report your answer to three digits after the decimal

Answer 1

Answer : The molar heat of solution for the salt is 452.9 kJ/mole

Explanation :

First we have to calculate the heat of solution.

$q=m\times c\times \Delta T$

where,

q = heat of solution = ?

c = specific heat of water = specific heat of solution = $4.184J/g.K=4.184J/g^oC$

m = mass of solution = 107.093 g

• Mass of solution = Mass of water + Mass of salt
• Mass of solution = 100.000 g + 7.093 g = 107.093 g

$\Delta T$ = change in temperature = $T_2-T_1=(80.000-61.128)=18.872^oC=$

Now put all the given values in the above formula, we get:

$q=107.093g\times 4.184J/g^oC\times 18.872^oC$

$q=8456.111J=8.456kJ$

Now we have to calculate the molar heat of solution for the salt, in kJ/mol.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = molar heat of solution = ?

q = heat required = 8.456 kJ

m = mass of salt = 7.093 g

Molar mass of salt = 379.984 g/mol

$\text{Moles of salt}=\frac{\text{Mass of salt}}{\text{Molar mass of salt}}=\frac{7.093g}{379.984g/mole}=0.01867mole$

$\Delta H=\frac{8.456kJ}{0.01867mole}=452.9kJ/mole$

Therefore, the molar heat of solution for the salt is 452.9 kJ/mole